CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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A macroscopic row term defeats averaging

Statement refuted

The assertion that independent centered rows automatically satisfy a weak law with normalization $n$ is false. Assume countable choice and dependent choice. A witness is the row of length $n$ defined by $X_{n, 1} = n ϵ_{n}$ and $X_{n, k} = 0$ for $2 \leq k \leq n$ , where the $ϵ_{n}$ are independent fair signs. With $S_{n} = \sum_{k = 1}^{n} X_{n, k}$ , $S_{n} / n$ does not converge in probability to zero.

Facts & Assumptions

[F1]

Chebyshev weak law for uncorrelated arrays: For each $n \geq 1$ , let $X_{n, 1}, \dots, X_{n, r_{n}}$ be square-integrable real random variables on one probability space, pairwise uncorrelated within the row, where $r_{n} \geq 0$ is finite. Set $S_{n} = \sum_{k = 1}^{r_{n}} X_{n, k}$ and let $b_{n} > 0$ be deterministic. If $v_{n} : = b_{n}^{- 2} \sum_{k = 1}^{r_{n}} Var (X_{n, k}) ⟶ 0,$ then $(S_{n} - E S_{n}) / b_{n} \to 0$ in $L^{2}$ and in probability. More precisely, its second moment is $v_{n}$ , and its probability of absolute value at least $ε > 0$ is at most $v_{n} / ε^{2}$ . No independence between rows is required.

[F2]

Convergence in probability: For real random variables $(X_{n})$ and $X$ on one probability space, write $X_{n} \to X$ in probability when, for every $ε > 0$ , $P (∣ X_{n} - X ∣ > ε) ⟶ 0.$ This is precisely def-convergence-in-measure for the probability measure.

[F3]

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure $ν$ on $(S, Σ)$ is the common law of a countable independent family of $S$ -valued random elements.

Counterexample

Given: The construction and assumptions above.

1.1

Under countable choice and dependent choice take IID fair signs using the countable-copy theorem. A row consisting of one random entry and constants is independent: any finite intersection of coordinate events reduces to the one nonconstant event or is empty. Each entry is centered and square-integrable. Its variance sum is $n^{2}$ , so the row weak law has normalized variance $1$ , not a quantity tending to zero.

F3F1givenalgebra

2.1

The sum is exactly $S_{n} = n ϵ_{n}$ . Thus $P (∣ S_{n} / n ∣ > 1 / 2) = 1$ for every $n \geq 1$ , contradicting the defining requirement for convergence in probability to zero. At $n = 1$ the row has only its one random entry. This is an array witness and asserts no failure of the IID integrable weak law.

F2step 1.1algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Theorem 2.2.6, p. 59, direct counterexample when its variance condition fails (standard reference, not scraped)